Q 58. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 53–58, find functions f and g so that $f\circ g=H$.

$H\left(x\right)=\left|2x^2+3\right|$

Step-by-Step Solution

Answer

The required functions are $f\left(x\right)=\left|x\right|; g\left(x\right)=2x+3$.

Step 1. Given information.

The given function is:

$H\left(x\right)=\left|2x^2+3\right|$

In the given function H takes $2x+3$and takes the absolute value of the function.

Now decompose H  by taking the absolute value of the function. 

Let's take $g\left(x\right)=2x+3$ and $f\left(x\right)=\left|x\right|$.

Step 2. Find f ∘ g .

$(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$

Substitute $g\left(x\right)=2x+3$ in the function $f\left(g\right(x\left)\right)$,

Then the function will become $f(2x+3)$.

Now replace x  with $2x+3$ in $f\left(x\right)=\left|x\right|$,

$$\begin{gathered} f(2x+3)=\left|2x+3\right| \\ \left|2x+3\right|=H\left(x\right) \end{gathered}$$

As we can see that $f\circ g=H$, therefore the values of the function that we assumed are correct.

$f\left(x\right)=\left|x\right|; g\left(x\right)=2x+3$.